GATE Overflow Test Series | Operating Systems | Test 2 | Question: 14

Question

The address sequence generated by tracing a particular program executing in a pure demand based paging system with $100$ records per page with $2$ free main memory frames and LRU replacement policy is recorded as follows. The number of page faults will be
$$0100, 0432, 0101, 0612, 0102, 0103, 0104, 0101, 0611, 0102, 0103, 0104, 0101$$

Comments

I’m getting 4 page faults.

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@anupam_yedida

Read the discussion on this question

https://gateoverflow.in/2619/Gate-cse-1995-question-2-7?show=367816#c367816

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Better strategy for this question would be to convert the given addresses to Page Numbers by dividing it by 100 e.g page number for 0100 would be 01 , similarly for 0432 it would be 04 and so on…

Now we got the Page Numbers and we have 2 Frames are available and hence we can go ahead with the algorithm.

Please feel free to correct me if I am wrong somewhere

Answer 1

$0100$ - $1$ page fault. Records $100-199$ in memory.

$0432$ - $2$ page faults. Records $400-499$ in memory.

$0101$ - page hit.

$0612$ - $3$ page faults. Records $600-699$ in memory. Records $400-499$ replaced.

$0102$ - page hit.

$0103$ - page hit.

$0104$ - page hit.

$0101$ - page hit.

$0611$ - page hit.

$0102, 0103, 0104, 0101$ - page hits.

So, total number of page faults $= 3.$